Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x-6y &= 3 \\ 6x+7y &= -5\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $7y = -6x-5$ Divide both sides by $7$ to isolate $y$ $y = {-\dfrac{6}{7}x - \dfrac{5}{7}}$ Substitute this expression for $y$ in the first equation. $9x-6({-\dfrac{6}{7}x - \dfrac{5}{7}}) = 3$ $9x + \dfrac{36}{7}x + \dfrac{30}{7} = 3$ Simplify by combining terms, then solve for $x$ $\dfrac{99}{7}x + \dfrac{30}{7} = 3$ $\dfrac{99}{7}x = -\dfrac{9}{7}$ $x = -\dfrac{1}{11}$ Substitute $-\dfrac{1}{11}$ for $x$ back into the top equation. $9( -\dfrac{1}{11})-6y = 3$ $-\dfrac{9}{11}-6y = 3$ $-6y = \dfrac{42}{11}$ $y = -\dfrac{7}{11}$ The solution is $\enspace x = -\dfrac{1}{11}, \enspace y = -\dfrac{7}{11}$.